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Let $K_1, K_2, \dots$ be a countable sequence of fields, and let $\prod_{\mathcal F} K_i$ be the ultraproduct with respect to some nonprincipal ultrafilter $\mathcal F$.

Question: Can there be a field isomorphism $\prod_{\mathcal F} K_i \cong \mathbb R$?

I think I gather that this can’t happen if each $K_i$ is isomorphic to $\mathbb R$ — you always get some infinitesimals in the ultrapower (but I don’t actually know why).

But maybe it can happen if the $K_i$ are smaller?

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    $\begingroup$ I don't know ultraproducts, but a product of two fields has zero divisors, so is not a field. $\endgroup$
    – Ben McKay
    Commented Mar 29 at 19:21
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    $\begingroup$ An ultraproduct of fields is a field. $\endgroup$
    – Dave Benson
    Commented Mar 29 at 19:23
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    $\begingroup$ @BenMcKay Yeah, ultraproducts are designed to get around this. In fact, by Łos’ theorem an ultraproduct of structures satisfies all the first-order statements which are true in all those structures (or even just those statements true in ultrafilter-large-many of the structures). For example, this is true for the formula $\forall x (x = 0 \vee \exists y (xy = 1))$. $\endgroup$
    – Tim Campion
    Commented Mar 29 at 19:25
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    $\begingroup$ Ultraproducts are also designed to be $\aleph_1$-saturated. $\endgroup$
    – YCor
    Commented Mar 30 at 21:47
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    $\begingroup$ @TimCampion: Below the downvote button on the question, there are two further buttons. The first is "Save this question." The second is "Show activity on this post." That shows useful information like what you want, and more, linking here: mathoverflow.net/posts/467999/timeline $\endgroup$
    – aorq
    Commented Mar 31 at 2:23

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The answer is no, because such ultrapowers are always $\aleph_1$-saturated, but $\mathbb{R}$ is not.

More concretely, the ultraproduct will be an ordered field with uncountable cofinality — every countable set is bounded above. To see this, observe first that because the order $x<y$ is definable in the real field, if this field were realized as an ultraproduct, then almost every $K_i$ will have to be an ordered field. It now follows that every countable sequence in the ultraproduct will be bounded, because like Hausdorff we can climb above any countable sequence of functions. Specifically, for any countably many $f_0,f_1,f_2,\ldots\in\prod_i K_i$, let $f(i)=\max_{k\leq i} f_k(i)+1$ in $K_i$, a finite supremum on each coordinate. The function $f$ is eventually above every $f_k$ on a tail, and so $[f]$ is above every $[f_k]$ in the ultraproduct.

But not every countable set in $\mathbb{R}$ is bounded, since for instance $\mathbb{Z}$ is unbounded in $\mathbb{R}$.

Incidentally, you asked about infinitesimals, and these exist in the ultraproduct by the same kind of reasoning. Just let $e(i)=1/i$ in $K_i$, and it follows that for any specific natural number $n$ we have $e(i)<1/n$ for almost all $i$, and so $[e]$ is infinitesimal in the ultraproduct. In fact, the positive elements of the ultraproduct have uncountable coinitiality, meaning that every countable set of positive elements is bounded below. Given $f_0,f_1,\ldots$ of positive elements, let $e(i)$ be the minimum of $f_k(i)$ for $k\leq i$, and then $0<[e]\leq[f_k]$ in the ultraproduct for every $k$.

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    $\begingroup$ Perfect, thanks! Apparently I have to wait a few minutes to accept your answer. $\endgroup$
    – Tim Campion
    Commented Mar 29 at 19:26
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    $\begingroup$ Hmm…. I think real-closed fields are not finitely axiomatizable, right? So I’d think you could take an ultraproduct of fields $K_n$ where $K_n$ has roots of polynomials of degree $\leq 2n-1$ but not all polynomials of degree $2n+1$ (so that $K_n$ is not real-closed) and in the ultraproduct get a real closed field. I think almost all $K_n$ have to at least be formally real though, you can axiomatize that finitely by saying that $x^2 + y^2$ always has a square root and is never zero unless $x = y = 0$. (Of course, this doesn’t change the answer to my question.) $\endgroup$
    – Tim Campion
    Commented Mar 29 at 19:37
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    $\begingroup$ Ah, that's right. But they'd have to be ordered, so this is enough for the argument. That is, we can define $x\leq y\iff\exists z( x+z^2=y)$, and this has to be an order in almost every $K_i$. $\endgroup$
    – Joel David Hamkins
    Commented Mar 29 at 19:38
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    $\begingroup$ Incidently, the infinitesimals come from the same idea, since the positive part of the field will have uncountable coinitiality. That is, any countable sequence of positive elements in the ultraproduct is bounded below by a positive element. Just take reciprocols and apply Hausdorff, or apply Hausdorff directly to get below on a tail. $\endgroup$
    – Joel David Hamkins
    Commented Mar 29 at 19:54
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    $\begingroup$ An analogous argument shows that $p$-adic fields (i.e. finite extensions of $\mathbf{Q}_p$) are not isomorphic to ultraproducts over countable sets. $\endgroup$
    – YCor
    Commented Apr 1 at 12:55

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